Transcript Brooker Chapter 5 - Volunteer State Community College

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Genetics: Analysis and Principles
Robert J. Brooker
CHAPTER 5
LINKAGE AND
GENETIC MAPPING
IN EUKARYOTES
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5.1 LINKAGE AND
CROSSING OVER

In eukaryotic species, each linear
chromosome contains a long piece of DNA


A typical chromosome contains many hundred
or even a few thousand different genes
The term linkage has two related meanings


1. Two or more genes can be located on the
same chromosome
2. Genes that are close together tend to be
transmitted as a unit
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5-3

Chromosomes are called linkage groups


They contain a group of genes that are linked together
The number of linkage groups is the number of
types of chromosomes of the species

For example, in humans




22 autosomal linkage groups
An X chromosome linkage group
A Y chromosome linkage group
Genes that are far apart on the same chromosome
may independently assort from each other

This is due to crossing-over
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5-4
Crossing Over May Produce
Recombinant Phenotypes

In diploid eukaryotic species, linkage can be
altered during meiosis as a result of crossing
over

Crossing over


Occurs during prophase I of meiosis at the
bivalent stage
Non-sister chromatids of homologous
chromosomes exchange DNA segments
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5-5
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B
A
B b
A a
B
A
b
a
Diploid cell after
chromosome replication
Diploid cell after
chromosome replication
Fig. 5.1(TE
Art)
Meiosis
B
A
b
a
b
a
Meiosis
B
A
B
A
B
a
b
a
b
a
b
A
Possible haploid cells
(a) Without crossing over, linked alleles
segregate together.
Possible haploid cells
(b) Crossing over can reassort linked
alleles.
These haploid cells contain a
combination of alleles NOT
found in the original
chromosomes
These are
termed
parental or
nonrecombinant
cells
Figure 5.1
This new combination of
alleles is a result of
genetic recombination
These are termed
nonparental or recombinant
cells
5-7
Bateson and Punnett Discovered Two
Traits That Did Not Assort Independently

In 1905, William Bateson and Reginald Punnett
conducted a cross in sweet pea involving two
different traits


Flower color and pollen shape
This is a dihybrid cross that is expected to yield a
9:3:3:1 phenotypic ratio in the F2 generation

However, Bateson and Punnett obtained surprising
results
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5-8
Figure 5.2
A much greater proportion
of the two types found in
the parental generation
5-9
Morgan Provided Evidence for the
Linkage of Several X-linked Genes



The first direct evidence of linkage came from
studies of Thomas Hunt Morgan
Morgan investigated several traits that followed an
X-linked pattern of inheritance
Figure 5.3 illustrates an experiment involving three
traits



Body color
Eye color
Wing length
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5-11
x
y+ w+ m+ Y
yy ww mm
F1 generation contains wild-type
females and yellow-bodied,
white-eyed, miniature-winged
males.
F1 generation
x
y+y w+w m+m
ywmY
P Males
P Females


Morgan observed a much higher proportion of the
combinations of traits found in the parental generation
Morgan’s explanation:
 All three genes are located on the X chromosome
 Therefore, they tend to be transmitted together as a unit
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5-13
Morgan Provided Evidence for the
Linkage of Several X-linked Genes

1. Why did the F2 generation have a
significant number of nonparental
combinations?

2. Why was there a quantitative difference
between the various nonparental
combinations?
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5-14
Let’s reorganize Morgan’s data by considering the pairs of
genes separately

Gray body, red eyes
1,159
Yellow body, white eyes
1,017
Gray body, white eyes
Yellow body, red eyes
Total
17
12
2,205
Red eyes, normal wings
770
White eyes, miniature wings
716
Red eyes, miniature wings
White eyes, normal wings
Total
401
318
2,205
But this nonparental
combination was rare
It was fairly common
to get this nonparental
combination
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5-15

Morgan made three important hypotheses to
explain his results

1. The genes for body color, eye color and wing
length are all located on the X-chromosome


2. Due to crossing over, the homologous X
chromosomes (in the female) can exchange
pieces of chromosomes


They tend to be inherited together
This created new combination of alleles
3. The likelihood of crossing over depends on
the distance between the two genes

Crossing over is more likely to occur between two
genes that are far apart from each other
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5-17
Figure 5-5
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 5.4
These parental phenotypes are
the most common offspring
These recombinant offspring
are not uncommon
because the genes are far apart
5-18
Figure 5.4
These recombinant offspring
are fairly uncommon
because the genes are very close together
These recombinant offspring
are very unlikely
1 out of 2,205
5-19
Chi Square Analysis

This method is frequently used to determine
if the outcome of a dihybrid cross is
consistent with linkage or independent
assortment
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5-20
Creighton and McClintock Experiment
C
wx
Normal
chromosome 9
Parental
chromosomes
c
Abnormal
chromosome 9
Knob
Interchanged
piece from
chromosome 8
(a) Normal and abnormal chromosome 9
C = Colored
c = colorless
Wx = Starchy endosperm
wx = waxy endosperm

Wx
Crossing over
c
wx
Nonparental
chromosomes
C
Wx
(b) Crossing over between normal and
abnormal chromosome 9
Figure 5.6
5-30
Interpreting the Data

Parent A
Parent B
C wx (nonrecombinant)
c Wx (nonrecombinant)
C Wx (recombinant)
c wx (recombinant)
c Wx
c wx
By combining these gametes into a Punnett square, the
following types of offspring can be produced
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5-32
Ambiguous phenotypes that
could be produced whether
or not recombination
occurred in parent A
So let’s start by
considering the
unambiguous
phenotypes
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5-33

The colored, waxy phenotype (Cc wxwx) can occur
only if



Recombination did not occur in parent A
AND
Parent A passed the knobbed, translocated chromosome
to its offspring
This was the case, as shown in the data table
below
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5-34

The colorless, waxy phenotype (cc wxwx) can
occur only if



Recombination did occur in parent A
AND
Parent A passed a chromosome 9 that had a
translocation but was knobless
This was the case, as shown in the data table
below
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5-35
The Data
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5-31

These observations were consistent with the
idea that a cross over occurred between the
C and wx genes

As stated by Creighton and McClintock:

“Pairing chromosomes, heteromorphic in two
regions, have been shown to exchange parts at
the same time they exchange genes assigned to
these regions.”
5-36
5.2 GENETIC MAPPING IN
PLANTS AND ANIMALS

Genetic mapping is also known as gene mapping
or chromosome mapping

Its purpose is to determine the linear order of
linked genes along the same chromosome

Figure 5.8 illustrates a simplified genetic linkage
map of Drosophila melanogaster
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5-42
Each gene has its
own unique locus
at a particular site
within a
chromosome
Figure 5.8
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5-43

Experimentally, the percentage of recombinant
offspring is correlated with the distance between the
two genes



If the genes are far apart  many recombinant offspring
If the genes are close  very few recombinant offspring
Map distance = Number of recombinant offspring X 100
Total number of offspring

The units of distance are called map units (mu)
 They are also referred to as centiMorgans (cM)

One map unit is equivalent to 1% recombination frequency
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5-45
Chromosomes are
the product of a
crossover during
meiosis in the
heterozygous parent
Recombinant
offspring are fewer
in number than
nonrecombinant
offspring
Figure 5.9
5-47


The data at the bottom of Figure 5.9 can be used to
estimate the distance between the two genes
Number of recombinant offspring X 100
Map distance =
Total number of offspring
=
76 + 75
542 + 537 + 76 + 75
X 100
= 12.3 map units
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5-48
Alfred Sturtevant’s Experiment

The first genetic map was constructed in 1911 by
Alfred Sturtevant


He was an undergraduate who spent time in the
laboratory of Thomas Hunt Morgan
Sturtevant wrote:

“In conversation with Morgan … I suddenly realized that
the variations in the length of linkage, already attributed
by Morgan to differences in the spatial orientation of the
genes, offered the possibility of determining sequences
[of different genes] in the linear dimension of the
chromosome. I went home and spent most of the night
(to the neglect of my undergraduate homework) in
producing the first chromosome map, which included the
sex-linked genes, y, w, v, m, and r, in the order and
approximately the relative spacing that they still appear
on the standard maps.”
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5-49
Figure 5.10
5-52
The Data
Alleles
Concerned
Number Recombinant/ Percent Recombinant
Total Number
Offspring
y and w/w-e
214/21,736
1.0
y and v
1,464/4,551
32.2
y and r
115/324
35.5
y and m
260/693
37.5
w/w-e and v
471/1,584
29.7
w/w-e and r
2,062/6,116
33.7
w/w-e and m
406/898
45.2
v and r
17/573
3.0
v and m
109/405
26.9
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5-53
Interpreting the Data

In some dihybrid crosses, the percentage of
nonparental (recombinant) offspring was rather low



For example, there’s only 1% recombinant offspring in
the crosses involving the y and w or w-e alleles
This suggests that these two genes are very close
together
Other dihybrid crosses showed a higher
percentage of nonparental offspring


For example, crosses between the v and m alleles
produced 26.9% recombinant offspring
This suggests that these two genes are farther apart
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5-54

Sturtevant assumed that the map distances would
be more accurate among genes that are closely
linked


Therefore, his map is based on the following distances
 y – w (1.0), w – v (29.7), v – r (3.0) and v – m (26.9)
Sturtevant also considered map distances amongst
gene pairs to deduce the order of genes




Percentage of crossovers between w and r was 33.7
Percentage of crossovers between w and v was 29.7
Percentage of crossovers between v and r was 3.0
Therefore, the gene order is w – v – r
 Where v is closer to r than it is to w
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5-55

Sturtevant collectively considered all these data
and proposed the following genetic map

Sturtevant began at the y gene and mapped the
genes from left to right
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5-56

A close look at Sturtevant’s data reveals two points
that do not agree very well with his genetic map



The y and m dihybrid cross yielded 37.5% recombinants
 But the map distance is 57.6
The w and m dihybrid cross yielded 45.2% recombinants
 But the map distance is 56.6
So what’s up?

As the percentage of recombinant offspring approaches a
value of 50 %
 This value becomes a progressively more inaccurate
measure of map distance
 Refer to Figure 5.11
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5-57
Figure 5.11

When the distance between two genes is large
 The likelihood of multiple crossovers increases
 This causes the observed number of recombinant offspring
to underestimate the distance between the two genes
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5-58
Figure 5-12a
Copyright © 2006 Pearson Prentice Hall, Inc.
Trihybrid Crosses
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5-59
Trihybrid Crosses


Data from trihybrid crosses can also yield information
about map distance and gene order
The following experiment outlines a common strategy for
using trihybrid crosses to map genes
 In this example, we will consider fruit flies that differ in
body color, eye color and wing shape






b = black body color
b+ = gray body color
pr = purple eye color
pr+ = red eye color
vg = vestigial wings
vg+ = normal wings
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5-59

Step 1: Cross two true-breeding strains that differ
with regard to three alleles.
Female is mutant
for all three traits

Male is homozygous
wildtype for all three
traits
The goal in this step is to obtain aF1 individuals that
are heterozygous for all three genes
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5-60

Step 2: Perform a testcross by mating F1 female
heterozygotes to male flies that are homozygous
recessive for all three alleles

During gametogenesis in the heterozygous female F1 flies,
crossovers may produce new combinations of the 3 alleles
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5-61

Step 3: Collect data for the F2 generation
Number of Observed Offspring
(males and females)
Phenotype
Gray body, red eyes, normal wings
411
Gray body, red eyes, vestigial wings
61
Gray body, purple eyes, normal wings
2
Gray body, purple eyes, vestigial wings
30
Black body, red eyes, normal wings
28
Black body, red eyes, vestigial wings
1
Black body, purple eyes, normal wings
60
Black body, purple eyes, vestigial wings
412
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5-62

The three genes exist as two alleles each
Therefore, there are 23 = 8 possible combinations of
F2 offspring
If the genes assorted independently, all eight
combinations would occur in equal proportions
It is obvious that they are far from equal

In the offspring of crosses involving linked genes,






Parental phenotypes occur most frequently
Double crossover phenotypes occur least frequently
Single crossover phenotypes occur with “intermediate”
frequency
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5-63

The combination of traits in the double crossover tells us
which gene is in the middle
 A double crossover separates the gene in the middle from
the other two genes at either end

In the double crossover categories, the recessive purple
eye color is separated from the other two recessive alleles
 Thus, the gene for eye color lies between the genes for
body color and wing shape
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5-64

Step 4: Calculate the map distance between pairs of
genes
 To do this, one strategy is to regroup the data
according to pairs of genes


From the parental generation, we know that the
dominant alleles are linked, as are the recessive alleles
This allows us to group pairs of genes into parental and
nonparental combinations



Parentals have a pair of dominant or a pair of recessive alleles
Nonparentals have one dominant and one recessive allele
The regrouped data will allow us to calculate the map
distance between the two genes
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5-65
Parental offspring
Gray body, red eyes
(411 + 61)
Black body, purple eyes
(412 + 60)
Total
Nonparental Offspring
Total
472
Gray body, purple eyes
(30 + 2)
32
472
Black body, red eyes
(28 + 1)
29
944

61
The map distance between body color and eye color is
61
X 100 = 6.1 map units
Map distance =
944 + 61
5-66
Parental offspring
Gray body, normal wings
(411 + 2)
Black body, vestigial wings
(412 + 1)
Total
Nonparental Offspring
Total
413
Gray body, vestigial wings
(30 + 61)
91
413
Black body, normal wings
(28 + 60)
88
826

179
The map distance between body color and wing shape is
179
X 100 = 17.8 map units
Map distance =
826 + 179
5-67
Parental offspring
Red eyes, normal wings
(411 + 28)
Purple eyes, vestigial wings
(412 + 30)
Total
Nonparental Offspring
Total
439
Red eyes, vestigial wings
(61 + 1)
62
442
Purple eyes, normal wings
(60 + 2)
62
881

124
The map distance between eye color and wing shape is
124
X 100 = 12.3 map units
Map distance =
881 + 124
5-68

Step 5: Construct the map

Based on the map unit calculation the body color and
wing shape genes are farthest apart
 The eye color gene is in the middle

The data is also consistent with the map being drawn
as vg – pr – b (from left to right)

In detailed genetic maps, the locations of genes are
mapped relative to the centromere
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5-69

To calculate map distance, we have gone
through a method that involved the separation of
data into pairs of genes (see step 4)

An alternative method does not require this
manipulation
 Rather, the trihybrid data is used directly

This method is described next
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5-70
Phenotype
Number of
Observed
Offspring
Gray body,
purple eyes,
vestigial wings
30
Black body,
red eyes,
normal wings
28
Gray body,
red eyes,
vestigial wings
61
Black body,
purple eyes,
normal wings
60
Gray body,
purple eyes,
normal wings
2
Black body,
red eyes,
vestigial wings
1
Single crossover
between b and pr
30 + 28
= 0.058
1,005
Single crossover
between pr and vg
61 + 60
Double crossover,
between b and pr,
and between
pr and vg
1+2
= 0.120
1,005
= 0.003
1,005
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5-71

To determine the map distance between the genes, we
need to consider both single and double crossovers

To calculate the distance between b and pr
 Map distance = (0.058 + 0.003) X 100 = 6.1 mu

To calculate the distance between pr and vg
 Map distance = (0.120 + 0.003) X 100 = 12.3 mu

To calculate the distance between b and vg


The double crossover frequency needs to be multiplied by two
 Because both crossovers are occurring between b and vg
Map distance = (0.058 + 0.120 + 2[0.003]) X 100
= 18.4 mu
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5-72

Alternatively, the distance between b and vg can be
obtained by simply adding the map distances between
b and pr, and between pr and vg
 Map distance = 6.1 + 12.3 = 18.4 mu

Note that in the first method (grouping in pairs), the
distance between b and vg was found to be 17.8 mu.
 This slightly lower value was a small underestimate
because the first method does not consider the double
crossovers in the calculation
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5-73
Figure 5-8
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 5-10
Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 5-11
Copyright © 2006 Pearson Prentice Hall, Inc.
Interference

The product rule allows us to predict the likelihood of a
double crossover from the individual probabilities of each
single crossover
P (double crossover) = P (single crossover X P (single crossover
between b and pr)
between pr and vg)
= 0.061 X 0.123 = 0.0075

Based on a total of 1,005 offspring
 The expected number of double crossover offspring is
= 1,005 X 0.0075 = 7.5
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5-74
Interference

Therefore, we would expect seven or eight offspring to be
produced as a result of a double crossover

However, the observed number was only three!



Two with gray bodies, purple eyes, and normal sings
One with black body, red eyes, and vestigial wings
This lower-than-expected value is due to a common genetic
phenomenon, termed positive interference
 The first crossover decreases the probability that a
second crossover will occur nearby
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5-75


Interference (I) is expressed as
 I = 1 – C
 where C is the coefficient of coincidence

Observed number of double crossovers
C=
Expected number of double crossovers

C=
3
7.5
= 0.40
I = 1 – C = 1 – 0.4
= 0.6 or 60%
 This means that 60% of the expected number of
crossovers did not occur
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5-76

Since I is positive, this interference is positive interference

Rarely, the outcome of a testcross yields a negative value
for interference
 This suggests that a first crossover enhances the rate of
a second crossover

The molecular mechanisms that cause interference are not
completely understood
 However, most organisms regulate the number of
crossovers so that very few occur per chromosome
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5-77
5.3 GENETIC MAPPING IN
HAPLOID EUKARYOTES

Much of our earliest understanding of genetic
recombination came from the genetic analyses of fungi

Fungi may be unicellular or multicellular organisms
They are typically haploid (1n)
They reproduce asexually and, in many cases, sexually



The sac fungi (ascomycetes) have been particularly useful
to geneticists because of their unique style of sexual
reproduction
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5-78
Meiosis produces
four haploid cells,
termed spores
These are
enclosed in a sac
termed an ascus
Figure 5.12
5-79



The cells of a tetrad or octad are contained within
a sac
In other words, the products of a single meiotic
division are contained within one sac
This is a key feature that dramatically differs from
sexual reproduction in animals and plants
 In animals, for example



Oogenesis only produces a single functional egg
Spermatogenesis produces sperm that are mixed
with millions of other sperm
Using a microscope, researchers can dissect asci
and study the traits of each haploid spore
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5-80
Types of Tetrads or Octads

The arrangement of spores within an ascus
varies from species to species

Unordered tetrads or octads


Ascus provides enough space for the spores to
randomly mix together
Ordered tetrads or octads

Ascus is very tight, thereby preventing spores from
randomly moving around
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5-81
Ascus provides
space for spores to
randomly mix
together
Tight ascus
prevents mixing
of spores
Mold
Yeast
Figure 5.13
Unicellular alga
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5-82
Ordered Tetrad Analysis

Ordered tetrads or octads have the following key
feature


The position and order of spores within the ascus is
determined by the divisions of meiosis and mitosis
This idea is schematically shown in Figure 5.13b

The example depicts ordered octad formation in
Neurospora crassa


Spores that carry the A allele show orange pigmentation
Spores that carry the a (albino) allele are white
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5-83
Pairs of daughter
cells are located
next to each other
Figure 5.13
All eight cells are
arranged in a linear,
ordered fashion
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020

The genetic content of spores in ordered
tetrads can be determined


This allows experimenters to map the distance
between a single gene and the centromere
The logic of this mapping technique is based
on the following features of meiosis


Centromeres of homologous chromosomes
separate during meiosis I
Centromeres of sister chromatids separate
during meiosis II
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This 4:4 arrangement of spores within
the ascus is termed a first-division
segregation (FDS) or an M1 pattern
Octad contains a linear arrangement of
4 haploid cells with the A allele which
are adjacent to 4 with the a allele
Because the A and a alleles have
segregated from each other after
meiosis I
Figure 5.14 (a) No crossing over
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These arrangement of
The A and a alleles do not
spores are termed a
second-division segregation segregate until meiosis II
(SDS) or M2 patterns
Figure 5.14 (b) Single crossing over
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
The percentage of M2 asci can be used to calculate
the map distance between the centromere and the
gene of interest
Figure 5.15
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
Therefore the chances of getting a 2:2:2:2 or 2:4:2 pattern
depend on the distance between the gene of interest and the
centromere

To calculate this distance, the experimenter must count the
number of SDS asci, as well as the total number of asci
 In SDS asci, only half of the spores are actually the
product of a crossover
 Therefore
(1/2) (Number of SDS asci) X 100
Map distance =
Total number of asci
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Unordered Tetrad Analysis

Unicellular algae
Figure 5-18
Copyright © 2006 Pearson Prentice Hall, Inc.
Unordered Tetrad Analysis

Unordered tetrads contain randomly arranged
groups of spores

An experimenter can do a dihybrid cross and then
determine the phenotypes of the spores
 Such an analysis can determine if two genes are
linked or assort independently
 It can also be used to compute distance
between two linked genes
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5-90
Unordered Tetrad Analysis

Consider a diploid yeast zygote with the genotype
ura+ura-2 arg+arg-3



ura+ and arg+ = Normal alleles required for uracil and
arginine biosynthesis, respectively
ura-2 and arg-3 = Defective alleles
 Result in strains that require uracil and arginine in
their growth medium
Figure 5.16 illustrates the assortment of the two
genes in the unordered tetrad
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Figure 5.16
PD ascus: contains
100% parental cells
T ascus: contains
50% parental cells and
50% recombinant cells
NPD ascus: contains
100% recombinant cells
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
If the two genes assort independently



The number of asci with a parental ditype is expected to
equal the number with a nonparental ditype
Thus, 50% recombinant spores are produced
If the two genes are linked

The type of crossover between them determines what
type of ascus is produced
 No crossovers yield the parental ditype
 Single crossovers produce the tetratype
 Double crossovers can yield any of the three types

The actual type produced depends on the combination of
chromatids that are involved
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Figure 5.17
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Figure 5.17
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
As in conventional mapping, the map distance is calculated
as the % of offspring that carry recombinant chromosomes
NPD + (1/2) (T)
Map distance =
X 100
Total number of asci

This calculation is fairly reliable over a short distance


However, over long distances it is not
 Because it does not adequately account for double crossovers
A more precise way to calculate map distance
Single crossover tetrads +
(2) (Double crossover tetrads)
Map distance =
Total number of asci
X 0.5 X 100
Crossover tetrads also contain 50%
nonrecombinant chromosomes
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

For the equation to be useful, it needs to be related to the
number of various types obtained by experimentation
So let’s take another look at Figure 5.17




The parental ditype (PD) and tetratype (T) are ambiguous
 They can each be derived in two different ways
The nonparental ditype (NPD), however, is unambiguous
 It can only be produced from a double crossover (DCO)
1/4 of all the double crossovers are nonparental ditypes
 Therefore, DCO = 4 X NPD
But what about single crossovers (SCO)?


Notice that T asci can result from SCO or DCO
Since there are two kinds of T that are due to DCO
 The actual number of T arising from DCO is 2NPD
 So, T = SCO + 2NPD
 Therefore, SCO = T – 2NPD
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
Now we have accurate measures of both SCO and DCO


SCO = T – 2NPD and DCO = 4NPD
So, let’s substitute these values into our previous equation
Single crossover tetrads +
(2) (Double crossover tetrads)
Map distance =
Map distance =
X 0.5 X 100
Total number of asci
(T – 2NPD) + (2) (4NPD)
Total number of asci
T + 6NPD
Map distance =
Total number of asci
X 0.5 X 100
X 0.5 X 100
A more accurate measure of map distance because the
equation considers both single- and double-crossovers
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